Supplemental Notes for Chapter 4
Second Law: Concepts and Applications
Real, Irreversible, Quasi-static, and Reversible
Partially quasi-static
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Quasi-static
Real
(Irreversible)
Reversible
Internally reversible
Quasi-static processes
- Along a quasi-static path all intermediate states are equilibrium states; thus
from postulate I quasi-static paths for closed, simple systems can be described by two
independent properties.
- From postulate II, if a system progressing along a quasi-static path is “isolated”
from its environment, then the values of all properties will remain constant and equal to
those just before the isolation.
- Quasi-static processes occur at finite rates but not so rapidly that the system is
able to adjust on a molecular level. There would not be, in general, gradients of any
intensive properties, such as temperature, pressure, density, etc.
- Expanding a gas contained in a frictionless piston (mass m, area a)-cylinder is
not a quasi-static process:
1. pull stops
Pa
2. rapid expansion during which a definite dP/dz
gradient exists in the gas phase
3.
piston
moves rapidly as Pgas is greater than
Pgas
mg
Pa +
a
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1
Without any friction present, the gas expansion will clearly not be quasi-static.
If friction is present so that the expansion process occurs very slowly, dP/dz would be
negligible and the properties of the gas would remain constant if the expansion process
were stopped – that is, the system would stay in some stable equilibrium state. Thus,
with friction present in this manner, the expansion process is quasi-static. A similar
situation is encountered in the gas cylinder blowdown. The valve controls the blowdown
rate, resulting in a quasi-static process for the gas contained in the cylinder.
In fact, the adiabatic tank blowdown process could be modeled as a closed system gas
expansion against a massless piston that is frictionally damped to keep Pgas = Poutside. In
this case, the process is quasi-static and:
d U = − PdV = Pd ( NV )
since N = constant
dU = NC v dT = − NPdV
for an ideal gas with P = RT/V, by eliminating V, we get
dT/T = – R/Cv (dV/V) = –R/Cv ( dT/T – dP/P)
(R/Cv+1)dT/T = (R/Cv)dP/P
If we had eliminated T, then dP/P =-(1+R/Cv) dV/V and the same equations result by
treating the system as open. Upon integration, we obtain the familiar relationships for a
reversible, adiabatic expansion (or compression) of an ideal gas, namely,
PVκ = constant or equivalently T/Ti = (P/Pi)R/Cp = (P/Pi)(Κ-1)/ Κ
where κ ≡ C p / Cv and C p = Cv + R
Reversible Processes
- Via Postulate II, if any (real or ideal) system in a non-equilibrium state is
isolated, it will tend toward a state of equilibrium.
- All real or natural processes are not reversible. Hence reversible processes are
only idealizations that are very useful in showing limiting behavior. The performance of
real processes is frequently compared with ideal performance under reversible conditions.
- "In a reversible process, all systems must be in states of equilibrium at all times,
that is all subsystems must traverse quasi-static paths."
- A system undergoing a reversible process is no more than differentially
removed from an equilibrium state – the system passes through a set of equilibrium
states.
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- "A process will be called reversible if a second process could be performed in
at least one way so that the system and all elements of its environment can be restored to
their respective initial states, except for differential changes of second order." For
example, in a reversible expansion or compression δ(δW) ≈ dPdV
- If a cyclic process A → B → A is reversible, then when the process is carried
out, no changes will occur in any other bodies. For example, if A → B involves the
absorption of a quantity of heat Q, then B → A will reject the same quantity Q to the
environment.
- Any reversible process is also quasi-static, but the reverse is not necessarily
true.
- Simple systems undergoing reversible processes have no internal gradients of
temperature or pressure.
- Friction and other dissipative forces are not present in reversible processes. A
truly reversible process will always require an infinitesimal driving force to ensure that
energy transfer occurs without degradation, hence its rate would be infinitely slow.
Therefore, a reversible process always can be shown to require a minimum a mount of
work or will yield a maximum amount of work.
- Heat engines in reversible processes operate at maximum efficiency
Summary of the 2nd Law
- The 1st law involves primarily the principle of energy conservation and is not
sufficient to describe how a natural process will proceed.
- The 2nd law is concerned with describing the direction is which a process can
take place. For example, the flow of heat from a hot to a cold body.
- The 2nd law describes, in mathematical terms, the physical impossibility of
reversing Joule's experiments. It is not possible to convert heat into an equivalent amount
of work – some amount of heat must be transferred to a second body or the environment
in the process of converting heat into work.
- Carnot heat engines operate cyclically and reversibly
between two isothermal reservoirs at TH and TC
and a work reservoir. The efficiency of the
Carnot process for converting heat into work is
= – δWc/δQH = (TH−TC)/TH TH>TC
so 0 < ηc 1.0
ηc
3
TH
δQH
δWc
δQC
TC
− Given a reversible process where temperature changes, it is always possible to find a
reversible zig-zag path consisting of adiabatic-isothermal-adiabatic steps such that the
heat interaction in the isothermal step is equal to the heat interaction of the original
process.
− Definition of entropy S as an derived state function
dS ≡ δQrev
/ T (an exact differential)
∆ S = ∫ d S and
∫dS = 0
− Clausius inequality for describing heat interactions between two isothermal reservoirs
at TA and TB.
δQ A / TA + δQB / TB ≥ 0
− For any fully reversible process, the equality applies, for all others the inequality
applies
− A reversible process adiabatic process occurs at constant entropy.
− For any real process;
(1) dS > 0 (system + surroundings)
(2) ∆ S universe
= ∆ S system + ∆ S surroundings > 0
− Entropy is a measure of the degradation of work producing potential.
− All natural processes in isolated, closed systems always occur in a direction
that increases entropy.
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Combined First and Second Laws
1. Closed, single phase, simple systems
From the 1st Law:
dE = dU = δQ + δW
For an internally reversible, quasi-static process with only PdV work:
δQ = δQrev = Td S and δW = δWrev = − PdV
Therefore, dU= TdS−PdV which also provides a way to determine entropy changes:
Basis: 1 mole of ideal gas of constant Cp
dS=dU/T + P/TdV
dS =
Cv
P
dT + dV
T
T
with PV=RT
dS =
Cv
P R
dP 
dT +  dT − RT 2 
T P
T
P 
combining terms:
dS =
Cv + R
dP
dT − R
T
P
or
2
T
P
∫ d S = ∆S = C p ln T21 − Rln P21
1
5
For ideal gas only
2. Open, single phase, simple systems
Insulation
For an internally reversible, quasi-static
process with one component entering and
leaving the system, all intensive properties
must remain the same. Hence,
�Wrev
�nin
� Surface
(E, V, S, T, P)
Tin = Tout = T
�Qrev
Pin = Pout = P
U in = U out = U
�nout
Sin = Sout = S
Adapted from Tester, J. W. and Modell, Michael. Thermodynamics and Its Applications.
Upper Saddle River, NJ: Prentice Hall PTR, 1997, p. 87.
Vin = Vout = V
Image by MIT OCW.
From a mole balance, dN = δnin − δnout . Now the 1st Law, which describes an energy
balance, can be written with only PdV work as:
d E = dU = δQrev − PdV + (U + PV )dN
and likewise an entropy balance can be formulated as:
d S = δQrev / T + SdN + δ S gen = δQrev / T + SdN
since δS gen = 0 for this reversible case. Now by substituting δQrev into the 1st Law
expression:
d U = Td S − PdV + (U + PV − TS )dN = Td S − PdV + µdN
where μ is the chemical potential defined as:
μ ≡ G ≡ U + PV − TS = H − TS
This result can be generalized for a multicomponent, single phase system that is
traversing a quasi-static path as,
6
n
d U = Td S − PdV + ∑ µ i dN i
i
where U is a continuous function of n+2 variables: U = f (S ,V , N i ) i=1,…n and
n
d U = (∂U / δ S )V , N d S + (∂U / ∂V )S , N dV + ∑(∂U / ∂N i )S ,V , N
i
j (i )
dN i
i
T
-P
Which is often referred to as the Fundamental Equation of Thermodynamics.
3. Availability (maximum and minimum work concepts)
As described in Section 14.1, consider a process
shown at the right that interacts with a work
reservoir and rejects heat to the surroundings.
Other constraints are:
�nin
�nout
Primary system
Small carnot engine
Work reservoir
�Qs
�Wc
Secondary system
- reversible/quasi-static operation
- steady state δn
≡ δnin = δnout
thus δN = 0 for primary system
- Carnot heat engine operates cyclicly
so δQs is at the system temperature T
- all heat δQR is rejected isothermally at T#
�Ws
�QR
Heat reservoir
at To
Adapted from Tester, J. W. and Modell, Michael. Thermodynamics and Its Applications. Upper
Saddle River, NJ: Prentice Hall PTR, 1997, p. 588.
Image by MIT OCW.
A differential steady state 1st Law balance around the process (primary system) yields:
d E = 0 = δQs + δWs + (H in − H out ) δn
(1)
Where δWs represents the shaft work contribution. A steady state 1st law balance around
the Carnot heat engine gives:
d E = 0 = −δQs + δQR + δWc
Where the – minus sign on δQs reflects its directional change relative to the primary
system, i.e., the engine is receiving heat from the system.
δQs = δQ R + δWc
(2)
Where δWc represents the Carnot heat engine work contribution. A steady state entropy
balance for the composite secondary system – the process and the heat engine yields:
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d S = 0 = + δQR / To + (S in − Sout ) δn
(3)
Combining equations (1), (2), and (3) to eliminate δQs and δQR and solving for the total
work interaction (Carnot + shaft work) gives with rearrangement:
∑ δWi = δWtotal = (δWc + δWs ) = [(H out − H in ) − To (Sout − Sin )]δn
(4)
Since we are dealing with a reversible process, Eq. (4) gives the maximum work per mole
that could be produced (or the minimum work required). This is called the availability or exergy:
B ≡ H − To S
∆B ≡ H out − H in − To (Sout − Sin ) = ∆H − To ∆S
(5)
Now Eq. (4) can be rewritten
δWmax = (δWc + δWs ) = ∆Bδn
(6)
or by taking the time derivative to give maximum power:
W max = (∆B ) δn / δt = (∆B ) n
Clearly, the availability is a state function in the strictest mathematical sense so the
maximum (or minimum) work associated with any steady state process is also
independent of the path.
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